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For oxalic acid, $\mathrm{K}_{\rm a1}> \mathrm{K}_{\rm a2}$.

My teacher told me that after first deprotonation, $\ce{COO-}$ acts as an electron donating group and hence reduces the acidic strength of $\ce{HC2O4-}$ (due to the +I effect of $\ce{COO-}$ on the second $\ce{COOH}$).

Why is $\ce{COO-}$ an electron donating group when $\ce{COOH}$ is an electron withdrawing group?

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  • $\begingroup$ 1.25, 4.14 according to en.wikipedia.org/wiki/Oxalic_acid suggests it may be even slightly withdrawing. $\endgroup$ – Mithoron Oct 6 '15 at 19:46
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    $\begingroup$ After loosing $\ce{H+}$ wont it become rich in electrons and act as an $\ce{+I}$ group ? $\endgroup$ – Sujith Sizon Oct 15 '15 at 12:22

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